I π 4 sec 2θ 1 dθ tan θ θπ 4 consequently i 1

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I=π/40(sec2θ1)dθ=tanθθπ/40.Consequently,I= 1π4.01510.0 pointsDetermine the integralI=1 + 4xx21dx .1.I= lnxx21+4x21+C2.I= 4 lnxx21+x21+C3.I= lnx+x21+ 4x21 +Ccorrect4.I= lnx+x214x21+C5.I= 4 lnx+x21x21+CExplanation:Letx= sec(θ) ; thendx= sec(θ) tan(θ)dθ,x21 = tan2(θ).In this caseI=1 + 4 sec(θ)tan(θ)sec(θ) tan(θ)dθ=(1 + 4 sec(θ)) sec(θ)dθ=(sec(θ) + 4 sec2(θ))dθ.Nowsec(θ)dθ= ln (|sec(θ) + tan(θ)|) +C1,whilesec2(θ) = tan(θ) +C2.ThusI= ln (|sec(θ) + tan(θ)|) + 4 tan(θ) +C .Consequently,I= lnx+x21+ 4x21 +CwithCan arbitrary constant.01610.0 points
cheatham (sc36975) – HW02 – um – (53890)9Evaluate the integralI=224x2x21dx .1.I=3 +22.I= 4(3 +2 )3.I= 4(32 )4.I= 2(3 +2 )5.I= 2(32 )correct6.I=32Explanation:Setx= sec(u). Thendx= sec(u) tan(u)du ,x21 = tan2(u),whilex=2=u=π4,x= 2=u=π3.In this case,I= 4π/3π/4sec(u) tan(u)sec2(u) tan(u)du=π/3π/44 cos(u)du= 4 sin(u)π/3π/4.Consequently,I= 2(32 ).01710.0 pointsEvaluate the integralI=31415 + 2xx2dx .1.I=4332.I=43π3.I=34.I=23πcorrect5.I=2336.I=πExplanation:By completing the square we see that15 + 2xx2= 16(x1)2,soI=31416(x1)2, dx .Now setx1 = 4 sin(u). Thendx= 4 cos(u)du ,whilex= 1=u= 0,x= 3=u=π6.ThusI= 4π/604 cos(u)4 cos(u)du=4uπ/60.Consequently,I=23π.01810.0 pointsEvaluate the integralI=324x24x+ 5dx .1.I=12π
cheatham (sc36975) – HW02 – um – (53890)102.I= 2π3.I= 14.I= 25.I=126.I=πcorrectExplanation:By completing the square we see thatx24x+ 5 = (x24x+ 4) + 54= (x2)2+ 1.In this caseI=324(x2)2+ 1.Now setx2 =u. For thendx=du, whilex= 2=u= 0,x= 3=u= 1,and soI= 41011 +u2du= 4 tan1u10.Consequently,I=π.

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